The equation of hyperbola $H$ is $\dfrac {(y-1)^{2}}{4}-\dfrac {(x-6)^{2}}{25} = 1$. What are the asymptotes?
Explanation: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac {(y-1)^{2}}{4} = 1 + \dfrac {(x-6)^{2}}{25}$ Multiply both sides of the equation by $4$ $(y-1)^{2} = { 4 + \dfrac{ (x-6)^{2} \cdot 4 }{25}}$ Take the square root of both sides. $\sqrt{(y-1)^{2}} = \pm \sqrt { 4 + \dfrac{ (x-6)^{2} \cdot 4 }{25}}$ $ y - 1 = \pm \sqrt { 4 + \dfrac{ (x-6)^{2} \cdot 4 }{25}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y - 1 \approx \pm \sqrt {\dfrac{ (x-6)^{2} \cdot 4 }{25}}$ $y - 1 \approx \pm \left(\dfrac{2 \cdot (x - 6)}{5}\right)$ Add $1$ to both sides and rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{2}{5}(x - 6)+ 1$